我试图从MySql数据库中获取值。我有一个表单,我想从MySql数据库中获取emailid。请帮忙
<tr><td><b>Email :</td><td><input type="text" name="txtemail" readonly value="<?php
$name=$_SESSION['usr_name'];
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="test"; // Database name
$conn=mysqli_connect($host,$username,$password) or die("cannot connect");
mysqli_select_db($conn,$db_name);
$result = mysqli_query($conn,"SELECT emailid FROM tabledetails WHERE
username='$name'");
echo $result;
mysqli_close($conn);
?>">
</td>
</tr>
答案 0 :(得分:1)
这在我的本地系统中完美运行:
根据您的要求尝试此代码:
<?php
$name=$_SESSION['usr_name']
$servername="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$dbname="test"; // Database name
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
else{
echo "Connected successfully";
}
$res = mysqli_query($conn,"SELECT emailid FROM tabledetails WHERE username='$name'");
while($row=mysqli_fetch_assoc($res))
{
$rows = $row['email'];
print_r $rows;
echo $rows;
}
?>
<tr><td><b>Email :</td><td><input type="text" name="txtemail" readonly value="<?php echo $rows ?>"/>
<强>输出: -
Connected successfully
Email :User
答案 1 :(得分:0)
删除此行。
mysqli_select_db($conn,$db_name);
并将第4个参数添加到mysqli_connect
$conn=mysqli_connect($host,$username,$password, $db_name) or die("cannot connect");
答案 2 :(得分:-1)
试试这个:
while($row = $result->fetch_assoc()) {
echo $row["Email"];
}